Question

YDSE with glass slab, Optical path
A flint glass and a crown glass are fitted on the two slits of a double slit apparatus. The thickeness of the strips is 0.40 mm and the separation between the slits is 0.12 cm. The refractive index of flint glass and crown glass are 1.62 and 1.52 resectively for the ligth of wavelength 480 nm . Which is used in the experiment.The interfrarance is observed on a screen a distance one meter away.
(a) What would be the fringe-whdth ?
(b) At what distance from the geometrical center will the nearest maximum be located?

Answer 1

(a) Fringe width is given by:

$\beta =\frac{\lambda D}{d}=\frac{480\times {10}^{-9}\times 0.12}{0.4}=144mm$

(b) Now the fringe shifts from the centre due to the upper slit is

$y_1=\frac{\beta }{\lambda }(\mu -1)t$

$y_1=\frac{100}{0.12}(1.62-1)0.4=206.66mm$

Now to fringe shifts from the centre due to the lower slit is

$y_2=\frac{\beta }{\lambda }(\mu -1)t$

$y_2=\frac{D}{d}(\mu -1)t$

$=\frac{100}{0.12}(1.52-1)0.4$

$=173.33mm$

So, net fringe shift

$y=206.66-173.33$

$y=33.33mm$